Martin Flashman's Courses
MATH 316 Real Analysis I Spring, 2016
Compactness and Uniform Continuity

Definition : Suppose `f: R -> R` and `S` is an interval. We say that `f` is uniformly continuous on `S` if 
for any `epsilon > 0`, there is a real number `delta >0` so that
for any `a` and `x` in `S`, if `|x-a| < delta` then `|f(x)-f(a)| < epsilon`.

Demonstration on of Uniform Continuity

Proposition: If `f` is uniformly continuous on and interval,`S`, then `f` is continuous on `S`.
Proof: Check the logic of the quantifiers for uniform continuity and continuity.

Example: `f(x) = 1/x` for `x \ne 0`, `f(0) = 0`; `S = (0, 1)`. `f` is continuous on `S`, but `f` is not uniformly continuous on `S`.
Example: `f(x) = mx + b` is uniformly continuous for `(-oo,oo)`.
Discussion: Suppose `epsilon > 0` is given. If `m=0` then let `delta = epsilon` will do. If `m \ne 0`, let `delta = epsilon/{|m|}`. The verification that for any `x` and `a` if  `|x-a| < delta` then `|f(x)-f(a)| < epsilon` is left as a routine exercise for the reader.

Theorem: A continuous function on a compact set of real numbers, `K` is uniformly continuous.
Plan- Given `epsilon >0`, find an open cover of `K` and use a finite subcover of `K` to find a`delta >0`
where if `|x-a| < delta` then `|f(x)-f(a)| < epsilon`.

For each `a in K`,
there is a `delta_a` where if  `|x-a| < delta_a`, then `|f(x) -f(a)| < epsilon/3`. [1]
Consider `N(a, delta_a/2)` and open set for each `a in K`. Then `O = {N(a,delta_a/2), a in K}` is an open cover of `K`.
Using `K` is compact, there is an open subcover determined by `{a_1, a_2, ..., a_n}` with related `delta_j= delta_{a_j}/2`.
Then `K sub N(a_1, delta_1) uu N(a_2,delta_2) uu ... uu N(a_n,delta_n)`.
`delta = min{delta_1, delta_2,...,delta_n} = 1/2 min{delta_{x_1},delta_{x_2},...,delta_{x_n}} `.

Now suppose  `x,a, in K` and `|x-a| < delta`.
Then for some `j` with `1  le j le n` we have `s in N(a_j,delta_j)` or `|a-a_j|<delta_j`.
Then `|x-a_j| le |x-a| + |a-a_j| < delta  + delta_j < delta_j  + delta_j =2 cdot delta_{a_j}/2 = delta _{a_j} `
Thus by [1]  `|f(x) - f(a_j) < epsilon/3` and `|f(a) -f(a_j)| <epsilon/3`.
So `|f(x) -f(a)| le |f(a)-f(a_j)| + |f(a_j) - f(x)| < 2 epsilon/3 < epsilon`.
Thus `f` is uniformly continuous on `K`.

For a well done proof with figures see

Theorem: Any continuous function on `[a,b]` is Darboux (or Riemann) integrable.
Suppose `f:[a,b] -> R` is continuous and `epsilon > 0`.
To show `f` is Darboux integrable we must find a partition `P` where `U(P,f) -L(P,f) < epsilon`.
Since `f` is continuous on `[a,b]` which is compact, `f` is uniformly continous on `[a,b]`. 
Choose `delta>0` so when `|x-t| < delta`,  `|f(x)-f(t)|< epsilon/{b-a}`.
Then by the extreme value theorem, for any interval, `I`, with length `l(I)<delta`, `lub{f(x): x in I} - glb{f(x):x in I} = f(c_M)-f(c_m) < epsilon/{b-a}.` 
Now choose N so that  `Delta x = {b-a)/N <delta`, `x_k = a + k Delta x`, and `P = {x_kP: k = 0,1, ..., n}`.

`U(P,f) -L(P,f) < \Sigma_1^N [epsilon/{b-a}] \ Delta x = [epsilon/{b-a}] \Sigma_1^N \ Delta x = [ epsilon/{b-a}] (b-a) = epsilon`

See also [An application of uniform continuity.]